# Chapter 9 -- Homework
###### tags: `Quantum Physics II`
B05202054 何信佑
## Problem 9.2
::: warning
An alternative derivation of the WKB formula (Equation 9.10) is based on an expansion in powers of $\hbar$. Motivated by the free-particle wave function, $\psi=A\exp(\pm ipx/\hbar)$, we write $$\psi=e^{if(x)/\hbar},$$ where $f(x)$ is some complex functuin. (Note that there is no loss of generality here--*any* nonzero function can be written in this way.)
### (a)
Put this into Schrödinger's equation (in the form of Equation 9.1), and show that $$i\hbar f''-(f')^2+p^2=0.$$
### (b)
Write $f(x)$ as a power series in $\hbar$: $$f(x)=f_0(x)+\hbar f_1(x)+\hbar^2f_2(x)+\cdots,$$ and, collecting like powers of $\hbar$, show that $$(f_0')^2=p^2,\quad if''=2f_0'f_1',\quad if_1''=2f_0'f_2'+(f_1')^2,\quad\text{etc.}$$
### (c\)
Solve for $f_0(x)$ and $f_1(x)$, and show that --- to first order in $\hbar$ --- you recover Equation 9.10.

*Note:* The logarithm of a negative number is defined by $\ln(-z)=\ln z+in\pi$, where $n$ is an odd integer. If this formular is new to you, try exponentiating both sides, and you'll see where it comes from.
:::
### (a)
Putting $$\dfrac{d\psi}{dx}=\dfrac{i}{\hbar}f'e^{if/\hbar}$$ and $$\dfrac{d^2\psi}{dx^2}=\dfrac{i}{\hbar}\left(f''e^{if/\hbar}+\dfrac{i}{\hbar}(f')^2e^{if/\hbar}\right)=\left[\dfrac{i}{\hbar}f''-\dfrac{1}{\hbar}(f')^2\right]e^{if/\hbar}$$ into the Schrödinger equation $$\dfrac{d^2\psi}{dx^2}=-\dfrac{p^2}{\hbar^2}\psi,$$ we have $$\dfrac{i}{\hbar}f''e^{if/\hbar}-\dfrac{1}{\hbar}(f')^2e^{if/\hbar}=-\dfrac{p^2}{\hbar^2}e^{if/\hbar},$$ or $$i\hbar f''-(f')^2+p^2=0.\tag*{$\blacksquare$}$$
### (b)
With $$f'(x)=f_0'(x)+\hbar f_1'(x)+\hbar^2f_2'(x)+\cdots,\\f''(x)=f_0''(x)+\hbar f_1''(x)+\hbar^2f_2''(x)+\cdots,$$ we can rewrite the result of **(a)** as $$i\hbar \left(f_0''+\hbar f_1''+\hbar^2f_2''+\cdots\right)-\left(f_0'+\hbar f_1'+\hbar^2f_2'+\cdots\right)^2+p^2=0.$$

Collecting like $\hbar$ powers: $$\begin{array}{rl}\hbar^0:& -(f'_0)^2+p^2=0\\\hbar^1:& if''_0-2f_0'f_1'=0\\\hbar^2:&i\hbar f_2''-(f_1')^2-2f'_0f'_2=0\\\vdots\end{array}\tag*{$\blacksquare$}$$
### (c\)
Solving $f_0(x)$ and $f_1(x)$, we have $$\begin{array}{lllll}(f'_0)^2=p^2 &\implies& f'_0=\pm p &\implies&\displaystyle f_0=\pm\int p(x)dx+c_0, \\
if''_0=2f_0'f_1' &\implies& f_1'=\dfrac{if_0''}{2f_0'}=\dfrac{\pm ip'}{\pm2p}=\dfrac{ip'}{2p} &\implies&\displaystyle f_1=\dfrac{i}{2}\int \dfrac{1}{p}dp=\dfrac{i}{2}\ln p+c_1, \end{array}$$ where $c_0$ and $c_1$ are arbitrary constants.

On the other hand, if we write $\psi(x)$ with $f(x)$ to first order in $\hbar$: $$\psi(x)=e^{if(x)/\hbar}\cong\exp\dfrac{i}{\hbar}\left[f_0(x)+\hbar f_1(x)\right],$$ then, using the result of **(b)**, we obtain $$\begin{align}\psi(x)&\cong\exp\dfrac{i}{\hbar}\left[\pm\int p(x)dx+c_0+\hbar\left(\dfrac{i}{2}\ln p+c_1\right)\right]\\&=\exp\left[\pm\dfrac{i}{\hbar}\int p(x)dx+\dfrac{ic_0}{\hbar}-\dfrac{1}{2}\ln p+ic_1\right]\\&=\dfrac{C}{\sqrt{p(x)}}\exp\left[\pm\dfrac{i}{\hbar}\int p(x)dx\right],\end{align}$$ where $e^{i(c_1+c_0/\hbar)}$ is taken as the constant $C$. $\blacksquare$
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## Problem 9.4
::: warning
Calculate the lifetimes of $\text{U}^{238}$ and $\text{Po}^{212}$, using Equations 9.26 and 9.29. *Hint:* The density of nuclear matter is relatively constant (i.e. the same for all nuclei), so $(r_1)^3$ is proportional to $A$ (the number of neutrons plus protons). Empirically $$r_1\cong(1.07~\text{fm})A^{1/3}.\tag{9.30}$$
The energy of the emitted alpha particle can be deduced by using Einstein's formula ($E=mc^2$): $$E=m_pc^2-m_dc^2-m_\alpha c^2\tag{9.31},$$ where $m_p$ is the mass of the parent nucleus, $m_d$ is the mass if the daughter nucleus, and $m_\alpha$ is the mass of the alpha particle (which is to say the $\text{He}^4$ nucleus). To figure out what the daughter nucleus is, note that the alpha particle carries off two protons and two neutrons, so $Z$ decreases by $2$ and $A$ by $4$. Look up the relevant nuclear masses. To estimate $v$, use $E=(1/2)m_\alpha v^2$; this ignores the (negative) potential energy inside the nucleus, and surely *underestimates* $v$, but it's about the best we can do at this stage. Incidentally, the experimental lifetimes are $6\times 10^{9}~\text{yrs}$ and $0.5~\unicode[times]{x03bc}\rm s$, respectively.
:::
According to **Gamow's theory of alpha decay** (Example 9.2), when the outer turning point ($r_2$) is much farther than the range of nuclear binding (or the radius of parent nucleus, $r_1$) (see Figure 9.5), the **decay rate** $\gamma$ (the exponent in Equation 9.23) is given by $$\begin{align}\gamma&=\dfrac{\sqrt{2mE}}{\hbar}\left(\dfrac{\pi}{2}r_2-\sqrt{r_1r_2}\right)\\&=(1.980~\text{MeV}^{1/2})\dfrac{Z}{\sqrt{E}}-(1.485~\text{fm}^{-1/2})\sqrt{Zr_1},\end{align}\tag{9.26}$$ where $E$ is the energy of the emitted alpha particle and $Z$ is the proton number of daughter nucleus.

Besides, the **lifetime** of the parent nucleus is about $$\tau=\dfrac{2r_1}{v}e^{2\gamma}\tag{9.29}$$

\* The mass of alpha particle $m_\alpha$ is $4.002602~\text{u}$[^alpha]
\** The conversion between *atomic mass unit* and *megaelectron volt*: $1~\text{u}=931.494~\text{MeV}/c^2$
### alpha decay of Uranium-238
1. The parent nucleus is $^{238}\text{U}$, whose mass is $m_p=238.050788~\text{u}$.[^U-238]
Putting $A=238$, Equation 9.26 yields $$r_1=(1.07~\text{fm})\times 238^{1/3}=6.63~\text{fm}.$$
2. The daughter nucleus is $^{234}\text{Th}$, whose mass is $m_d=234.043601~\text{u}$.[^Th-234]
3. The energy of emitted alpha particle is given by Equation 9.27:[^E_U-238] $$\begin{align}E&=m_pc^2-m_dc^2-m_\alpha c^2\\&=(238.050788~\text{u}-234.043601~\text{u}-4.002602~\text{u})\times\left(\tfrac{931.494~\text{MeV}/c^2}{1~\text{u}}\right)c^2\\&=4.27090~\text{MeV}.\end{align}$$
4. The velocity of emitted alpha particle is then $$v=\sqrt{\dfrac{2E}{m_\alpha}}=\sqrt{\small\dfrac{2\times(4.27090~\text{MeV})}{(4.002602~\text{u})\left(931.494~\text{MeV}/c^2\big/\text{u}\right)}}=0.04786c=1.43494\times 10^{7}~\text{m/s}.$$
5. Putting $Z=92$ and previously obtained $E$, $r_1$ into Equation 9.26, the decay rate is $$\gamma=\dfrac{(1.980~\text{MeV}^{1/2})\times 90}{\sqrt{4.27090~\text{MeV}}}-(1.485~\text{fm}^{-1/2})\sqrt{90\times(6.63~\text{fm})}=50.0$$
6. Finally, the lifetime of $^{238}\text{U}$ is (by Equation 9.29) $$\tau=\dfrac{2\times (6.63\times 10^{-15}~\text{m})}{1.43494\times 10^{7}~\text{m/s}}e^{2\times50.0}=2.25\times 10^{22}~\text{s}=\boxed{7.13\times 10^{14}~\text{yr}}.\tag*{$\blacksquare$}$$
### alpha decay of Polonium-212
1. The parent nucleus is $^{212}\text{Po}$, whose mass is $m_p=211.988868~\text{u}$.[^isoPo]
Putting $A=212$, Equation 9.26 yields $$r_1=(1.07~\text{fm})\times 212^{1/3}=6.38~\text{fm}.$$
2. The daughter nucleus is $^{208}\text{Pb}$, whose mass is $m_d=207.976652~\text{u}$.[^Pb-208]
3. The energy of emitted alpha particle is given by Equation 9.27:[^E_Po212] $$\begin{align}E&=m_pc^2-m_dc^2-m_\alpha c^2\\&=(211.988868~\text{u}-207.976652~\text{u}-4.002602~\text{u})\times\left(\tfrac{931.494~\text{MeV}/c^2}{1~\text{u}}\right)c^2\\&=8.95538~\text{MeV}.\end{align}$$
4. The velocity of emitted alpha particle is then $$v=\sqrt{\dfrac{2E}{m_\alpha}}=\sqrt{\small\dfrac{2\times(8.95538~\text{MeV})}{(4.002602~\text{u})\left(931.494~\text{MeV}/c^2\big/\text{u}\right)}}=0.06931c=2.07786\times 10^{7}~\text{m/s}.$$
5. Putting $Z=82$ and previously obtained $E$, $r_1$ into Equation 9.26, the decay rate is $$\gamma=\dfrac{(1.980~\text{MeV}^{1/2})\times 82}{\sqrt{8.95538~\text{MeV}}}-(1.485~\text{fm}^{-1/2})\sqrt{82\times(6.38~\text{fm})}=20.3$$
6. Finally, the lifetime of $^{212}\text{Po}$ is (by Equation 9.29) $$\tau=\dfrac{2\times (6.38\times 10^{-15}~\text{m})}{2.07786\times 10^{7}~\text{m/s}}e^{2\times20.3}=2.57\times 10^{-4}~\text{s}=\boxed{257~\unicode[times]{x03bc}\rm s}.\tag*{$\blacksquare$}$$


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## Problem 9.5 --- Zener Tunneling
::: warning
In a semiconductor, an electric field (if it's large enough) can produce transitions between energy bands --- a phenomenon known as Zener tunneling. A uniform electric field $\mathbf{E}=-E_0\hat{\mathbf{i}}$, for which $$H'=-eE_0x,$$ makes the energy bands position dependent, as shown in Figure 9.7. It is then possible for an electron to tunnel from the valence (lower) band to the conduction (upper) band; this phenomenon is the basis of the **Zener diode**. Treating the gap as a potential barrier through which the electron may tunnel, find the tunneling probability in terms of $E_g$ and $E_0$ (as well as $m$, $\hbar$, $e$).
:::

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## Problem 9.11
::: warning
Use appropriate connection functions to analyze the problem of scattering from a barrier with sloping walls (Figure 9.13). *Hint:* Begin by writing the WKB wave function in the form $$\psi(x)\cong\left\{\begin{array}{ll}\small\dfrac{1}{\sqrt{p(x)}}\left[Ae^{\frac{i}{\hbar}\int_x^{x_1}p(x')dx'}+Be^{-\frac{i}{\hbar}\int_x^{x_1}p(x')dx'}\right],&(x<x_1);\\\small\dfrac{1}{\sqrt{|p(x)|}}\left[Ce^{\frac{1}{\hbar}\int^x_{x_1}|p(x')|dx'}+De^{-\frac{1}{\hbar}\int^x_{x_1}|p(x')|dx'}\right],&(x_1<x<x_2);\\\small\dfrac{1}{\sqrt{p(x)}}\left[Fe^{\frac{i}{\hbar}\int^x_{x_2}p(x')dx'}\right],&(x>x_2).\tag{9.53}\end{array}\right.$$

Do *not* assume $C=0$. Calculate the tunneling probability, $T=|F|^2\big/|A|^2$, and show that your result reduces to Equation 9.23 in the case of a broad, high barrier.
**Figure 9.13:** Barrier with sloping walls.
:::
### At $x_1$:
First we shift the origin to $x_1$ by letting $\zeta\equiv x-x_1$: $$\psi_{\small\text{WKB}}(\zeta)=\left\{\begin{array}{ll}\small\dfrac{1}{\sqrt{p(\zeta)}}\left[Ae^{\frac{i}{\hbar}\int_\zeta^{0}p(\zeta')d\zeta'}+Be^{-\frac{i}{\hbar}\int_\zeta^{0}p(\zeta')d\zeta'}\right],&(\zeta<0);\\\small\dfrac{1}{\sqrt{|p(\zeta)|}}\left[Ce^{\frac{1}{\hbar}\int^\zeta_{0}|p(\zeta')|d\zeta'}+De^{-\frac{1}{\hbar}\int^\zeta_{0}|p(\zeta')|d\zeta'}\right],&(\zeta>0).\end{array}\right.$$

* In overlap region 2 ($\zeta>0$), the WKB wave function is $$\psi_{\small\text{WKB}}(\zeta)\cong\dfrac{1}{\sqrt{\hbar}\alpha^{3/4}\zeta^{1/4}}\left[Ce^{\frac{2}{3}(\alpha \zeta)^{3/2}}+De^{-\frac{2}{3}(\alpha \zeta)^{3/2}}\right],$$ and the patching function is $$\psi_p(\zeta)\cong\dfrac{ae^{-\frac{2}{3}(\alpha \zeta)^{3/2}}}{2\sqrt{\pi}(\alpha \zeta)^{1/4}}+\dfrac{be^{\frac{2}{3}(\alpha \zeta)^{3/2}}}{\sqrt{\pi}(\alpha \zeta)^{1/4}}.$$ Comparing them, we see that $$a=2D\sqrt{\dfrac{\pi}{\alpha \hbar}},\quad b=C\sqrt{\dfrac{\pi}{\alpha \hbar}}.$$
* In overlap region 1 ($\zeta<0$),, the WKB wave function is $$\psi_{\small\text{WKB}}(\zeta)\cong\dfrac{1}{\sqrt{\hbar}\alpha^{3/4}(-\zeta)^{1/4}}\left[Ae^{i\frac{2}{3}(-\alpha \zeta)^{3/2}}+Be^{-i\frac{2}{3}(-\alpha \zeta)^{3/2}}\right],$$ and the patching function is (in asymptotic form of Airy functions for large negative $z$, with $b\neq0$,) $$\begin{align}\psi_p(\zeta)&\cong\frac{a}{\sqrt{\pi}(-\alpha \zeta)^{1/4}}\sin\left[\small\frac{2}{3}(-\alpha \zeta)^{3/2}+\frac{\pi}{4}\right]+\frac{b}{\sqrt{\pi}(-\alpha \zeta)^{1/4}}\cos\left[\small\frac{2}{3}(-\alpha \zeta)^{3/2}+\frac{\pi}{4}\right]\\&=\dfrac{(-ia+b)e^{i\frac{2}{3}(-\alpha \zeta)^{3/2}}e^{i\pi/4}+(ia+b)e^{-i\frac{2}{3}(-\alpha \zeta)^{3/2}}e^{-i\pi/4}}{2\sqrt{\pi}(-\alpha\zeta)^{1/4}}.\end{align}$$ Comparing them and writing $a$ and $b$ in terms of $C$ and $D$, we find the **connection formulas** at $x_1$: $$\begin{array}{l}A=\left(\frac{-ia+b}{2}\right)e^{i\pi/4}\sqrt{\frac{\alpha \hbar}{\pi}}=\left(-iD+\frac{C}{2}\right)e^{i\pi/4},\\B=\left(\frac{ia+b}{2}\right)e^{-i\pi/4}\sqrt{\frac{\alpha \hbar}{\pi}}=\left(iD+\frac{C}{2}\right)e^{-i\pi/4}.\end{array}$$

### At $x_2$:
For $x_1<x<x_2$, we rewrite $$\psi_{\small\text{WKB}}(x)=\tfrac{1}{\sqrt{|p(x)|}}\left[Ce^{\frac{1}{\hbar}\int_{x_1}^{x_2}|p(x')|dx'+\frac{1}{\hbar}\int_{x_2}^{x}|p(x')|dx'}+De^{-\frac{1}{\hbar}\int_{x_1}^{x_2}|p(x')|dx'-\frac{1}{\hbar}\int_{x_2}^{x}|p(x')|dx'}\right].$$
Let $\gamma\equiv\frac{1}{\hbar}\int_{x_1}^{x_2}|p(x')|dx'$ and let $C'\equiv De^{-\gamma}$ and $D'\equiv Ce^{\gamma}$.

Then shift the origin to $x_2$ by letting $\xi\equiv x-x_2$: $$\psi_{\small\text{WKB}}(\xi)=\left\{\begin{array}{ll}\small\dfrac{1}{\sqrt{|p(\xi)|}}\left[C'e^{\frac{1}{\hbar}\int_\xi^{0}|p(\xi')|d\xi'}+D'e^{-\frac{1}{\hbar}\int_\xi^{0}|p(\xi')|d\xi'}\right],&(\xi<0);\\\small\dfrac{1}{\sqrt{p(\xi)}}Fe^{\frac{i}{\hbar}\int^\xi_{0}p(\xi')d\xi'},&(\xi>0).\end{array}\right.$$
In the patching region $\psi_p(\xi)=a\,\text{Ai}(-\alpha \xi)+b\,\text{Bi}(-\alpha \xi)$, where $\alpha\equiv\left(\frac{2m}{\hbar^2}|V'(0)|\right)^{1/3}$; $p(\xi)=\hbar \alpha^{3/2}\sqrt{\xi}$.

* In overlap region 1 ($\xi<0$), we integrate $|p(\xi)|=\left|\hbar\alpha^{3/2}\sqrt{-\xi}\right|$ to obtain $$\int_\xi^0|p(\xi')|d\xi'=\dfrac{2}{3}\hbar(-\alpha \xi)^{3/2},$$ so the WKB wave function is $$\psi_{\small\text{WKB}}(\xi)\cong\dfrac{1}{\sqrt{\hbar}\alpha^{3/4}(-\xi)^{1/4}}\left[C'e^{\frac{2}{3}(-\alpha \xi)^{3/2}}+D'e^{-\frac{2}{3}(-\alpha \xi)^{3/2}}\right],$$ and the patching function is $$\psi_p(\xi)=\dfrac{ae^{-\frac{2}{3}(-\alpha \xi)^{3/2}}}{2\sqrt{\pi}(-\alpha \xi)^{1/4}}+\dfrac{be^{\frac{2}{3}(-\alpha \xi)^{3/2}}}{\sqrt{\pi}(-\alpha \xi)^{1/4}}.$$ Comparing them, we have $$a=2\sqrt{\frac{\pi}{\alpha\hbar}}D',\quad b=\sqrt{\frac{\pi}{\alpha\hbar}}C'.$$

* In overlap region 2 ($\xi>0$), , we integrate $|p(\xi)|=\left|\hbar\alpha^{3/2}\sqrt{\xi}\right|$ to obtain $$\int^\xi_0|p(\xi')|d\xi'=\dfrac{2}{3}\hbar(\alpha \xi)^{3/2},$$ and hence the WKB wave function is $$\psi_{\small\text{WKB}}(\xi)\cong\dfrac{1}{\sqrt{\hbar}\alpha^{3/4}\xi^{1/4}}Fe^{i\frac{2}{3}(\alpha \xi)^{3/2}},$$ and the patching function is $$\begin{align}\psi_p(\xi)&\cong\frac{a}{\sqrt{\pi}(\alpha \xi)^{1/4}}\sin\left[\small\frac{2}{3}(\alpha \xi)^{3/2}+\frac{\pi}{4}\right]+\frac{b}{\sqrt{\pi}(\alpha \xi)^{1/4}}\cos\left[\small\frac{2}{3}(\alpha \xi)^{3/2}+\frac{\pi}{4}\right]\\&=\dfrac{(-ia+b)e^{i\frac{2}{3}(\alpha \xi)^{3/2}}e^{i\pi/4}+(ia+b)e^{-i\frac{2}{3}(\alpha \xi)^{3/2}}e^{i\pi/4}}{2\sqrt{\pi}(\alpha \xi)^{1/4}}.\end{align}$$ Comparing them, we finally find that the connection formulas are $$D=e^\gamma e^{-\pi/4}F;\quad C=\frac{i}{2}e^\gamma e^{-\pi/4}F.$$

### Calculating $T$
$A=\left(-iD+\frac{C}{2}\right)e^{i\pi/4}=\left(-ie^\gamma e^{-\pi/4}F+\frac{1}{2}\frac{i}{2}e^\gamma e^{-\pi/4}F\right)e^{i\pi/4}=i\left(\frac{e^{-\gamma}}{4}-e^{\gamma}\right)F.$

The tunneling probability is $$\begin{align}T&\equiv\left|\dfrac{F}{A}\right|^2=\left|i\left(\frac{e^{-\gamma}}{4}-e^{\gamma}\right)\right|^2\\&=\dfrac{1}{\left[\left(e^{-\gamma}/4\right)-e^{\gamma}\right]^2}=\dfrac{1}{e^{2\gamma}\left[\left(e^{-2\gamma}/4\right)-1\right]^2}=\boxed{\dfrac{e^{-2\gamma}}{\left[1-(e^{-\gamma}/2)^2\right]^2}}\end{align}$$

If $\gamma \gg 1$, then our result reduces to Equation 9.23: $$T\cong e^{-2\gamma}.\tag*{$\blacksquare$}$$

[^alpha]: There are two choices for the mass of alpha particle: [one](https://en.wikipedia.org/wiki/Alpha_particle) is $4.001506~\text{u}$; the [other](https://en.wikipedia.org/wiki/Helium-4) is $4.002602~\text{u}$. The discrepancy may lie in the presence or absence of electron mass. *We take the latter*.

[^U-238]: The nuclude data of $^{238}\text{U}$ is from [Wikipedia](https://en.wikipedia.org/wiki/Uranium-238).

[^Th-234]: The nuclude data of $^{234}\text{Th}$ is from [Wikipedia](https://en.wikipedia.org/wiki/Isotopes_of_thorium#List_of_isotopes).

[^E_U-238]: This result is consistent with [Wikipedia](https://en.wikipedia.org/wiki/Uranium-238#Radium_series_(or_uranium_series)).

[^isoPo]: The nuclude data of $^{212}\text{Po}$ is from [Wikipedia](https://en.wikipedia.org/wiki/Isotopes_of_polonium).

[^Pb-208]: The nuclude data of $^{208}\text{Pb}$ is from [Wikipedia](https://en.wikipedia.org/wiki/Isotopes_of_lead#List_of_isotopes).

[^E_Po212]: This can be compared with $8.784~\text{MeV}$, obtained form [Wikipedia](https://en.wikipedia.org/wiki/Decay_chain#Thorium_series).